# Why does Mascolell define second-order stochastic dominance as such?

Is not Mascolell's definition in his microeconomic theory of the second-order stochastic dominance narrower? He defines for distribution functions with the same mean only. Although he gives some motivation for doing this, somehow I do not get his motivation.

• @AlecosPapadopoulos If lottery $F$ fosds lottery $G$, then $E[F] \ge E[G]$. However, it is not true that whenever $E[F] \ge E[G]$, $F$ fosds $G$. For a counter-example, see Wikipedia. Hence, first-order stochastic dominance is more than just about the mean, and it is not equivalent to say that the dominant has a higher mean than the dominated. – Theoretical Economist Oct 21 '17 at 13:33
• @TheoreticalEconomist Thanks for spotting this. Indeed I had in mind expected utility "returns", $E[u(F)] \ge E[u(G)]$ for non-decreasing $u$, but I forgot the "utility" part and resulted in a mistaken assertion. – Alecos Papadopoulos Oct 21 '17 at 16:12